Решите уравнение: √(3х+1)+√(х-1)=6

Решите задачу:

$\sqrt{3x+1}+\sqrt{x-1} = 6, \\ \left \{ {{3x+1 \geq 0,} \atop {x-1 \geq 0;}} \right. \left \{ {{x \geq -\frac{1}{3},} \atop {x \geq 1;}} \right. \\ x \geq 1;$
$\sqrt{3x+1} = 6-\sqrt{x-1}, \\ (\sqrt{3x+1})^2 = (6-\sqrt{x-1})^2, \\ 3x+1 = 36-12\sqrt{x-1}+x-1, \\ 12\sqrt{x-1} = 34-2x, \\ (12\sqrt{x-1})^2 = (34-2x)^2, \\ 144(x-1) = 1156-136x+4x^2, \\ 4x^2-280x+1300 = 0, \\ x^2-70x+325 = 0, \\ D_{/4} = (-35)^2-1\cdot325 = 1225-325 = 900=30^2, \\ x_1=35-30=5, \\ x_2=35+30=65.$